jmc

Let $G$ be the foot of the perpendicular from $O$ to $AB$. Denote $x=EG$ and $y=FG$, and $x>y$ (since $AE<BF$ and $AG=BG$). Then $\tan \angle EOG=\frac{x}{450}$, and $\tan \angle FOG=\frac{y}{450}$. By the tangent addition rule $\left(\tan (a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}\right)$, we see that$$\tan 45=\tan (EOG+FOG)=\frac{\frac{x}{450}+\frac{y}{450}}{1-\frac{x}{450}\cdot \frac{y}{450}}.$$Since $\tan 45=1$, this simplifies to $1-\frac{xy}{450^2}=\frac{x+y}{450}$. We know that $x+y=400$, so we can substitute this to find that $1-\frac{xy}{450^2}=\frac{8}{9}⟹xy=150^2$. Substituting $x=400-y$ again, we know have $xy=(400-y)y=150^2$. This is a quadratic with roots $200\pm 50\sqrt{7}$. Since $y<x$, use the smaller root, $200-50\sqrt{7}$. Now, $BF=BG-FG=450-(200-50\sqrt{7})=250+50\sqrt{7}$. The answer is $250+50+7=\boxed{307}$.